Boundary Harnack Principle and Martin Boundary at Infinity for Subordinate Brownian Motions

Transcription

1 Potential Anal DOI /s Boundary Harnack Principle and Martin Boundary at Infinity for Subordinate Brownian Motions Panki Kim Renming Song Zoran Vondraček Received: 23 January 2013 / Accepted: 19 September 2013 Springer Science+Business Media Dordrecht 2013 Abstract In this paper we study the Martin boundary of unbounded open sets at infinity for a large class of subordinate Brownian motions. We first prove that, for such subordinate Brownian motions, the uniform boundary Harnack principle at infinity holds for arbitrary unbounded open sets. Then we introduce the notion of κ-fatness at infinity for open sets and show that the Martin boundary at infinity of any such open set consists of exactly one point and that point is a minimal Martin boundary point. Keywords Lévy processes Subordinate Brownian motion Harmonic functions Boundary Harnack principle Martin kernel Martin boundary Poisson kernel Mathematics Subject Classifications (2010 Primary 60J45; Secondary 60J25 60J50 P. Kim was supported by the National Research Foundation of Korea (NRF grant funded by the Korea government(mest ( R. Song was supported in part by a grant from the Simons Foundation ( Z. Vondraček was supported in part by the MZOS grant P. Kim Department of Mathematical Sciences and Research Institute of Mathematics, Seoul National University, Building 27, 1 Gwanak-ro, Gwanak-gu Seoul , Republic of Korea pkim@snu.ac.kr R. Song Department of Mathematics, University of Illinois, Urbana, IL 61801, USA rsong@math.uiuc.edu Z. Vondraček (B Department of Mathematics, University of Zagreb, Zagreb, Croatia vondra@math.hr

2 P. Kim et al. 1 Introduction and Main Results The study of the boundary Harnack principle for non-local operators started in the late 1990 s with 2 which proved that the boundary Harnack principle holds for the fractional Laplacian (or equivalently the rotationally invariant stable process in bounded Lipschitz domains. This boundary Harnack principle was extended to arbitrary open sets in 21. The final word in the case of the rotationally invariant α-stable process was given in 4 where the so called uniform boundary Harnack principle was proved in arbitrary open sets with a constant not depending on the set itself. Subsequently, the boundary Harnack principle was extended to more general symmetric Lévy processes, more precisely to subordinate Brownian motions with ever more weaker assumptions on the Laplace exponents of the subordinators, see 9, 11, 12 and8. For a boundary Harnack principle for subordinate Brownian motions with Gaussian components, see 13. Recently in 5, a boundary Harnack principle was established in the setting of jump processes in metric measure spaces. Let us be more specific and state the (slightly stronger assumptions under which the boundary Harnack principle was proved in 12. Let S = (S t t 0 be a subordinator (a nonnegative Lévy process with S 0 = 0 with Laplace exponent φ and W = (W t, P x t 0,x R d be a Brownian motion in R d, d 1, independent of S with E x e iξ (W t W 0 = e t ξ 2 ξ R d, t > 0. The process X = (X t, P x t 0,x R d defined by X t := W(S t is called a subordinate Brownian motion. It is a rotationally invariant Lévy process in R d with characteristic exponent φ( ξ 2 and infinitesimal generator φ(.here denotes the Laplacian and φ( is defined through functional calculus. The function φ is a Bernstein function having the representation φ(λ = a + bλ + (1 e λt μ(dt (0, where a, b 0 and μ is the measure satisfying (0, (1 tμ(dt <, called the Lévy measure of φ (or S. Recall that φ is a complete Bernstein function if the measure μ has a completely monotone density. For basic facts about complete Bernstein functions, see 19. Let us introduce the following upper and lower scaling conditions on φ at infinity: (H1: There exist constants 0 <δ 1 δ 2 < 1 and a 1, a 2 > 0 such that a 1 λ δ1 φ(t φ(λt a 2 λ δ2 φ(t, λ 1, t 1. (1.1 This is a condition on the asymptotic behavior of φ at infinity and it governs the behavior of the subordinator S for small time and small space (see 12, 14. Note that it follows from the second inequality above that φ has no drift, i.e., b = 0. Suppose that φ is a complete Bernstein function with the killing term a = 0 and that (H1 holds. The following boundary Harnack principle is proved in 12, Theorem 1.1: There exists a constant c = c(φ, d >0 such that for every z R d, every open set D R d, every r (0, 1 and any nonnegative functions u,v on R d which are regular harmonic in D B(z, r with respect to X and vanish in D c B(z, r, u(x v(x c u(y v(y, x, y D B(z, r/2.

3 Boundary Harnack Principle and Martin Boundary at Infinity Here, and in the sequel, B(z, r denotes the open ball in R d centered at z with radius r. This result was obtained as a simple consequence of the following approximate factorization of nonnegative harmonic functions, see 12, Lemma 5.5: There exists a constant c = c(φ, d >1 such that for every z R d, every open set D B(z, r and any nonnegative function u on R d which is regular harmonic in D with respect to X and vanishes a.e. in D c B(z, r, c 1 E x τ D B(z,r/2 c j( y z u(y dy u(x c E x τ D j( y z u(ydy (1.2 B(z,r/2 c for every x D B(z, r/2. Here w j( w denotes the density of the Lévy measure of X and τ D the first exit time of X from D. In the case of the rotationally invariant α-stable process, Eq. 1.2 is proved earlier in 4. Note that the boundary Harnack principle is a result about the decay of nonnegative harmonic functions near the (finite boundary points. It is an interesting problem to study the decay of non-negative harmonic functions at infinity (which may be regarded as a boundary point at infinity of unbounded sets. This is the main topic of the current paper. In order to study the behavior of harmonic functions at infinity, one needs large space and large time properties of the underlying process X. This requires a different type of assumption than (H1 which gives only small space and small time properties of X. Therefore, in addition to (H1, we will assume the corresponding upper and lower scaling conditions of φ near zero: (H2: There exist constants 0 <δ 3 δ 4 < 1 and a 3, a 4 > 0 such that a 3 λ δ4 φ(t φ(λt a 4 λ δ3 φ(t, λ 1, t 1. (1.3 This is a condition on the asymptotic behavior of φ at zero and it governs the behavior of the subordinator S for large time and large space (see 14 for details and examples. Using the tables at the end of 19, one can construct a lot of explicit examples of complete Bernstein functions satisfying both (H1 and (H2. Here area few of them: (1 φ(λ = λ α + λ β, 0 <α<β<1; (2 φ(λ = (λ + λ α β, α, β (0, 1; (3 φ(λ = λ α (log(1 + λ β, α (0, 1, β (0, 1 α; (4 φ(λ = λ α (log(1 + λ β, α (0, 1, β (0,α; (5 φ(λ = (log(cosh( λ α, α (0, 1; (6 φ(λ = (log(sinh( λ log λ α, α (0, 1. In the recent paper 14 we studied the potential theory of subordinate Brownian motions under the assumption that φ is a complete Bernstein function satisfying both conditions (H1 and (H2. We were able to extend many potential-theoretic results that were proved under (H1 (or similar assumptions on the small time and small space behavior for radii r (0, 1 to the case of all r > 0 (with a uniform constant not

4 P. Kim et al. depending on r > 0. In particular, we proved a uniform boundary Harnack principle with explicit decay rate (in open sets satisfying the interior and the exterior ball conditions which is valid for all r > 0. The current paper is a continuation of 14 and is based on the results of 14. For any open set D,weuseX D to denote the subprocess of X killed upon exiting D. IncaseD is a Greenian open set in R d we will use G D (x, y to denote the Green function of X D. For a Greenian open set D R d,let K D (x, y := G D (x, z j( z y dz, D (x, y D D c be the Poisson kernel of X in D D c. The first goal of this paper is to prove the following approximate factorization of regular harmonic functions vanishing at infinity. Theorem 1.1 Suppose that φ is a complete Bernstein function satisfying (H1 (H2, let d > 2(δ 2 δ 4, and let X be a rotationally invariant Lévy process in R d with characteristic exponent φ( ξ 2. For every a > 1, there exists C 1 = C 1 (φ, a >1 such that for any r 1, any open set U B(0, r c and any nonnegative function u on R d which is regular harmonic with respect to X in U and vanishes a.e. on B(0, r c \ U, it holds that C 1 1 K U(x, 0 u(z dz u(x C 1 K U (x, 0 u(z dz, (1.4 B(0,ar B(0,ar for all x U B(0, ar c. Note that K U (x, 0 = U G U(x, y j( y dy. A consequence of the assumption d > 2(δ 2 δ 4 (always true for d 2 in Theorem 1.1 is that the process X is transient and points are polar. Under this assumption, the Green function G(x, y of the process X exists, and by Eq. 2.9 below we have G U (x, y G(x, y x y d φ( x y 2 1. This will be used several times in this paper. In the case of the rotationally invariant α-stable process, Theorem 1.1 (for a = 2 was obtained in 16, Corollary 3 from Eq. 1.2 by using the inversion with respect to spheres and the Kelvin transform of harmonic functions for the stable process. Since the Kelvin transform method works only for stable processes we had to use a different approach to prove Eq We followed the method used in 12 toprove Eq. 1.2, making necessary changes at each step. The main technical difficulty of the proof is the delicate upper estimate of the Poisson kernel K B(0,r c(x, 0 of the complement of the ball given in Lemma 3.2, where the full power of the results from 14 was used. Theorem 1.1 gives the following scale invariant boundary Harnack inequality at infinity. Corollary 1.2 (Boundary Harnack Principle at Infinity Suppose that φ is a complete Bernstein function satisfying (H1 (H2, d> 2(δ 2 δ 4, and that X is a rotationally invariant Lévy process in R d with characteristic exponent φ( ξ 2. For each a > 1 there

5 Boundary Harnack Principle and Martin Boundary at Infinity exists C 2 = C 2 (φ, a >1 such that for any r 1, any open set U B(0, r c and any nonnegative functions u and v on R d that are regular harmonic in U with respect to X and vanish a.e. on B(0, r c \ U, it holds that C 1 u(y 2 v(y u(x v(x C u(y 2 v(y, for all x, y U B(0, arc. (1.5 The boundary Harnack principle is the main tool in identifying the (minimal Martin boundary (with respect to the process X of an open set. Recall that for κ (0, 1/2, anopensetd is said to be κ-fat open at Q D, if there exists R > 0 such that for each r (0, R there exists a point A r (Q satisfying B(A r (Q, κr D B(Q, r.ifd is κ-fat at each boundary point Q D with the same R > 0, D is called κ-fat with characteristics (R,κ.In case X is a subordinate Brownian motion via a subordinator with a complete Bernstein Laplace exponent regularly varying at infinity with index in (0, 1, itisshownin9 that the minimal Martin boundary of a bounded κ-fat open set can be identified with the Euclidean boundary. Corollary 1.2 enables us to identify the Martin boundary and the minimal Martin boundary at infinity of a large class of open sets with respect to X. Tobemore precise, let us first define κ-fatness at infinity. Definition 1.3 Let κ (0, 1/2. We say that an open set D in R d is κ-fat at infinity if there exists R > 0 such that for every r R, there exists A r R d such that B(A r,κr D B(0, r c and A r <κ 1 r. The pair (R,κ will be called the characteristics of the κ-fat open set D at infinity. Note that all half-space-like open sets, all exterior open sets and all infinite cones are κ-fat at infinity. Examples of disconnected open sets which are κ-fat at infinity are (i {x = (x 1,...,x d 1, x d R d : x d < 0 or x d > 1}; (ii n=1 B(x(n, 2 n 2 with x (n =2 n. Let D R d be an open set which is κ-fat at infinity with characteristics (R,κ.Fix x 0 D and define M D (x, y := G D(x, y G D (x 0, y, x, y D, y = x 0. As the process X D satisfies Hypothesis (B in 15, D has a Martin boundary M D with respect to X and M D (x, can be continuously extended to M D. A point w M D is called an infinite Martin boundary point if every sequence (y n n 1, y n D, converging to w in the Martin topology is unbounded (in the Euclidean metric. The set of all infinite Martin boundary points will be denoted by M D and we call this set the Martin boundary at infinity. By using the boundary Harnack principle at infinity we first show that if D is κ-fat at infinity, then there exists the limit M D (x, = lim M D(x, y. (1.6 y D, y

6 P. Kim et al. The existence of this limit shows that M D consists of a single point which we denote by. Finally, we prove that is a minimal Martin boundary point. These findings are summarized in the second main result of the paper. Theorem 1.4 Suppose that φ is a complete Bernstein function satisfying (H1 (H2, d > 2(δ 2 δ 4, and X is a rotationally invariant Lévy process in R d with characteristic exponent φ( ξ 2. Then the Martin boundary at inf inity with respect to X of any open set D which is κ-fat at inf inity consists of exactly one point. This point is a minimal Martin boundary point. We emphasize that this result is proved without any assumption on the finite boundary points. In particular, we do not assume that D is κ-fat. To the best of our knowledge, the only case where the Martin boundary at infinity has been identified is the case of the rotationally invariant α-stable process, see 4. Again, the Kelvin transform method was used to transfer results for finite boundary points to the infinite boundary point. As we have already pointed out, the Kelvin transform is not available for more general processes. We remark here that for one-dimensional Lévy processes (satisfying much weaker assumptions than ours it is proved in 20, Theorem 4 that the minimal Martin boundary at infinity for the half-line D = (0, is one point. The question of the Martin boundary at infinity is not addressed in 20. The paper is organized as follows. In the next section we introduce necessary notation and definitions, and recall some results that follow from (H1 and (H2 obtained in 14. Section 3 is devoted to the proofs of Theorem 1.1 and Corollary 1.2. At the end of the section we collect some consequences of these two results. In the first part of Section 4 we study non-negative harmonic functions in unbounded sets that are κ-fat at infinity. The main technical result is the oscillation reduction in Lemma 4.7 immediately leading to Eq Next we look at the Martin and the minimal Martin boundary at infinity and give a proof of Theorem 1.4. We finish the paper by discussing the Martin boundary of the half-space. Throughout this paper, the constants C 1, C 2, C 3,... will be fixed. The lowercase constants c 1, c 2,... will denote generic constants whose exact values are not important and can change from one appearance to another. The dependence of the lower case constants on the dimension d and the function φ may not be mentioned explicitly. The constant c that depends on the parameters δ i and a i, i = 1, 2, 3, 4, appearing in (H1 and (H2 will be simply denoted as c = c(φ. Wewilluse := to denote a definition, which is read as is defined to be. For a, b R, a b := min{a, b} and a b := max{a, b}. For any open set U, wedenotebyδ U (x the distance between x and the complement of U, i.e., δ U (x = dist(x, U c. For functions f and g, the notation f g means that there exist constants c 2 c 1 > 0 such that c 1 g f c 2 g. For every function f, we extend its definition to the cemetery point by setting f ( = 0. For every function f,let f + := f 0. We will use dx to denote the Lebesgue measure in R d and, for a Borel set A R d,wealsouse A to denote its Lebesgue measure. We denote B(x, r c := {y R d : x y > r}. Finally, for a point x = (x 1,...,x d 1, x d R d we sometimes write x = ( x, x d with x = (x 1,...,x d 1 R d 1.

7 Boundary Harnack Principle and Martin Boundary at Infinity 2 Preliminaries In this section we recall some results from 14. Recall that a function φ : (0, (0, is a Bernstein function if it is C function on (0, and ( 1 n 1 φ (n 0 for all n 1. It is well known that, if φ is a Bernstein function, then φ(λt λφ(t for all λ 1, t > 0. (2.1 Clearly Eq. 2.1 implies the following observation. Lemma 2.1 If φ is a Bernstein function, then every λ>0, 1 λ φ(λt 1 λ, for all t > 0. φ(t Note that with this Lemma, we can replace expressions of the type φ(λt, when φ is a Bernstein function, with λ>0 fixed and t > 0 arbitrary, with φ(t up to a multiplicative constant depending on λ. We will often do this without explicitly mentioning it. Recall that a subordinator S = (S t t 0 is simply a nonnegative Lévy process with S 0 = 0.LetS = (S t t 0 be a subordinator with Laplace exponent φ. The function φ is a Bernstein function with φ(0 = 0 so it has the representation ( φ(λ = bλ + 1 e λt μ(dt, (0, where b 0 is the drift and μ the Lévy measure of S. A Bernstein function φ is a complete Bernstein function if its Lévy measure μ has a completely monotone density, which will be denoted by μ(t. Throughout this paper we assume that φ is a complete Bernstein function. In this case, the potential measure U of S admits a completely monotone density u(t (cf. 19. Conditions (H1 (H2 imply that ( R δ1 δ 3 c 1 φ(r ( R δ2 δ 4 r φ(r c, 0 < r < R <. (2.2 r (See 14 for details. Using Eq. 2.2, we have the following result which will be used many times later in the paper. (See the proof of 12, Lemma 4.1 for similar computations. Lemma Assume (H1 and (H2. There exists a constant c = c(φ 1 such that λ 1 λ 1 λ 2 rφ(r 2 dr + 0 c 1 φ(λ λ 1 0 φ(r 2 1/2 dr cλ 1 φ(λ 2 1/2, for all λ>0, (2.3 λ 1 r 1 φ(r 2 dr cφ(λ 2, for all λ>0, (2.4 r 1 φ(r 2 1 dr cφ(λ 2 1, for all λ>0. (2.5

8 P. Kim et al. Recall that S = (S t t 0 is a subordinator with Laplace exponent φ. LetW = (W t t 0 be a d-dimensional Brownian motion, d 1, independent of S and with transition density q(t, x, y = (4πt d/2 e x y 2 4t, x, y R d, t > 0. The process X = (X t t 0 defined by X t := W(S t is called a subordinate Brownian motion. X is a rotationally invariant Lévy process with characteristic exponent φ( ξ 2, ξ R d. Throughout this paper X is always such a subordinate Brownian motion. The Lévy measure of X has a density J(x = j( x where j : (0, (0, is given by j(r := 0 (4πt d/2 e r2 /(4t μ(t dt. Note that j is continuous and decreasing. Recall that the infinitesimal generator L of the process X (e.g. 18, Theorem 31.5 is given by ( L f (x = f (x + y f (x y f(x1{ y ε} J(ydy (2.6 R d for every ε>0 and f C 2 b (Rd,whereC 2 b (Rd is the collection of C 2 functions which, along with its partial derivatives of up to order 2, are bounded. By the Chung-Fuchs criterion the process X is transient if and only if 1 0 λ d/2 1 φ(λ dλ<. It follows that X is always transient when d 3. Incase(H2 holds, X is transient provided δ 4 < d/2 (which is true if d 2. When X is transient the occupation measure of X admits a density G(x, y which is called the Green function of X and is given by the formula G(x, y = g( x y where g(r := 0 (4πt d/2 e r2 /(4t u(t dt. (2.7 Here u is the potential density of the subordinator S. Note that by the transience assumption, the integral converges. Moreover, g is continuous and decreasing. Furthermore, (H1 (H2 imply the following estimates. Theorem Assume both (H1 and (H2. (a It holds that j(r r d φ(r 2, for all r > 0. (2.8 (b If d > 2(δ 2 δ 4 then the process X is transient and it holds g(r r d φ(r 2 1, for all r > 0. (2.9

9 Boundary Harnack Principle and Martin Boundary at Infinity As a consequence of Eq. 2.8, we have Corollary 2.4 Assume (H1 and (H2. For every L > 1, there exists a constant c = c(l >0 such that j(r cj(lr, r > 0. (2.10 For any open set D, weuseτ D to denote the first exit time of D, i.e., τ D = inf{t > 0 : X t / D}. Given an open set D R d,wedefinext D (ω = X t (ω if t <τ D (ω and Xt D (ω = if t τ D (ω, where is a cemetery state. Let p(t, x, y be the transition density of X. By the strong Markov property, p D (t, x, y := p(t, x, y E x p(t τd, X τd, y ; t >τ D, x, y D, is the transition density of X D. A subset D of R d is said to be Greenian (for X if X D is transient. For a Greenian set D R d,letg D (x, y denote the Green function of X D, i.e., G D (x, y := 0 p D (t, x, ydt for x, y D. We define the Poisson kernel K D (x, z of D with respect to X by K D (x, z = G D (x, yj(y, z dy, (x, z D D c. (2.11 D Then by 7, Theorem 1 we get that for every Greenian open subset D, every nonnegative Borel measurable function f 0 and x D, E x f (XτD ; X τd = X τd = D c K D(x, y f (ydy. (2.12 Using the continuities of G D and J, one can easily check that K D is continuous on D D c. Equations 2.8 and 2.11 give the following estimates on the Poisson kernel of B(x 0, r for all r > 0. Proposition Assume (H1 and (H2. Thereexistc 1 = c 1 (φ > 0 and c 2 = c 2 (φ > 0 such that for every r > 0 and x 0 R d, K B(x0,r(x, y c 1 j( y x 0 r ( φ(r 2 φ((r x x 0 2 1/2 (2.13 for all (x, y B(x 0, r B(x 0, r c and c 1 j( y x 0 rφ(r 2 1 (2.14 K B(x0,r(x 0, y c 2 j( y x 0 φ(r 2 1, for all y B(x 0, r c. (2.15 To discuss the Harnack inequality and the boundary Harnack principle, we first recall the definition of harmonic functions.

10 P. Kim et al. Definition 2.6 A function f : R d 0, is said to be (1 harmonic in an open set D R d with respect to X if for every open set B whose closure is a compact subset of D, f (x = E x f (X(τB for every x B; (2.16 (2 regular harmonic in D for X if for each x D, f (x = E x f (X(τD ; τ D < ; (3 harmonic for X D if it is harmonic for X in D and vanishes outside D. We note that, by the strong Markov property of X, every regular harmonic function is automatically harmonic. Under the assumptions (H1 and (H2, the following uniform Harnack inequality and uniform boundary Harnack principle for all r > 0 are true. Theorem Assume (H1 and (H2. There exists c = c(φ > 0 such that, for any r > 0, x 0 R d, and any function u which is nonnegative on R d and harmonic with respect to X in B(x 0, r, we have u(x cu(y, for all x, y B(x 0, r/2. Theorem Assume (H1 and (H2. There exists a constant c = c(φ > 0 such that for every z 0 R d, every open set D R d,everyr> 0 and any nonnegative functions u,v in R d which are regular harmonic in D B(z 0, r with respect to X and vanish in D c B(z 0, r, we have u(x v(x c u(y v(y, for all x, y D B(z 0, r/2. For x R d,letδ D (x denote the Euclidean distance between x and D. Recall that δ D (x is the Euclidean distance between x and D c. In the next result we will assume that D satisfies the following two types of ball conditions with radius R: (i uniform interior ball condition: for every x D with δ D (x <R there exists z x D so that x z x =δ D (x and B(x 0, R D, x 0 := z x + R x z x x z x ; (ii uniform exterior ball condition: D is equal to the interior of D and for every y R d \ D with δ D (y <R there exists z y D so that y z y =δ D (y and B(y 0, R R d \ D, y 0 := z y + R y z y y z y. The following is the one of main results in 14 the global uniform boundary Harnack principle with explicit decay rate on open sets in R d with the interior and exterior ball conditions with radius R for all R > 0.

11 Boundary Harnack Principle and Martin Boundary at Infinity Theorem Assume (H1 and (H2. There exists c = c(φ > 0 such that for every open set D satisfying the interior and exterior ball conditions with radius R > 0, r (0, R,everyQ D and every nonnegative function u in R d whichisharmonicin D B(Q, r with respect to X and vanishes continuously on D c B(Q, r, we have u(x u(y c φ(δ D (y 2 φ(δ D (x 2 ( for every x, y D B Q, r. ( Boundary Harnack Principle at Infinity The goal of this section is to prove the scale invariant boundary Harnack principle at infinity (Theorem 1.1 and Corollary 1.2. In the remainder of this paper we assume that φ is a complete Bernstein function satisfying (H1 (H2 and d > 2(δ 2 δ 4,and that X is a rotationally invariant Lévy process in R d with characteristic exponent φ( ξ 2. Under these assumptions, by Eq. 2.9, g satisfies the following property which we will use frequently: For every L > 1, there exists c = c(l,φ>0 such that g(r cg(lr, r > 0. (3.1 To prove Theorem 1.1 we need several Lemmas. For x R d and 0 < r 1 < r 2,we use A(x, r 1, r 2 to denote the annulus {y R d : r 1 < y x r 2 }. Lemma 3.1 For every a (1,, there exists c = c(φ, a >0 such that for any r > 0 and any open set D B(0, r c we have P x (X τd B(0, r cr d K D (x, 0, x D B(0, ar c. Proof Let ψ Cc (Rd be a function such that 0 ψ 1, { 0, y > a + 1, ψ(y = 2 1, y 1, and sup d y R d i, j=1 2 y i y j ψ(y c 1 = c 1 (a. For r > 0 define ψ r (y := ψ(y/r. Then ψ Cc (Rd, 0 ψ r 1, { 0, y > a + 1 ψ r (y = 2 r, 1, y r, and sup d y R d i, j=1 2 y i y j ψ r (y c 1 r 2. Let x D B(0, ar c. Recall that L denotes the infinitesimal generator of X and is given by Eq Sinceψ r (x = 0 and D B(0, r c, by Dynkin s formula (see, for instance, 6, (5.8 we have E x ψr (X τd = G D (x, zlψ r (z dz D = G D (x, zlψ r (z dz + G D (x, zlψ r (z dz. D A(0,r,(a+2r D B(0,(a+2r c (3.2

12 P. Kim et al. For z D A(0, r,(a + 2r we have ( Lψ r (z = ψr (z + y ψ r (z ψ r (z y1 { y r} j( y dy R d ψ r (z + y ψ r (z ψ r (z y j( y dy + 2 c 2 r 2 { y r} y 2 j( y dy + 2 j( y dy { y r} {r< y } r r 2 tφ(t 2 dt + t 1 φ(t 2 dt, c 3 ( 0 r {r< y } j( y dy where in the last line we have used Eq Thus by using Eq. 2.4, we get Lψ r (z c 4 φ(r 2. By Lemma 2.1 we see that φ(r 2 and φ((a r 2 are comparable (with a constant depending on φ and a. Therefore Lψ r (z c 4 φ( z 2 c 5 z d j( z c 6 r d j( z, r < z <(a + 2r. (3.3 Now assume that z D B(0,(a + 2r c.thenψ r (z = 0 and ψ r (z = 0 (note that ψ r is zero in a neighborhood of z. Therefore ( Lψ r (z = ψr (z + y ψ r (z ψ r (z y1 { y r} j( y dy R d = ψ r (z + y j( y dy = ψ r (z + y j( y dy, R d { z+y a+1 2 r} where the last equality follows from the fact that ψ r (z + y = 0 only if z + y a+1 2 r. Suppose that z + y a+1 r. By the triangle inequality, 2 y z a r > z a a + 2 z = a + 3 z z 2(a It follows from Eq that j( y j( z /2 c 7 j( z. Thisimpliesthat Lψ r (z c 7 j( z dy c 8 r d j( z. (3.4 z+y a+1 2 r Combining Eqs we obtain E x ψr (X τd = G D (x, zlψ r (z dz + G D (x, zlψ r (z dz D A(0,r,(a+2r D B(0,(a+2r c c 9 r d G D (x, z j( z dz = c 9 r d K D (x, 0. D Finally, since 1 B(0,r ψ r, P x (X τd B(0, r E x ψr (X τd c 9 r d K D (x, 0. Lemma 3.2 Let 1 < p < q <. There exists c = c(φ, p, q >1 such that for all r 1/4 it holds that for all x A(0, pr, qr and z B(0, r. K B(0,r c(x, z cr d φ(r 2 1/2 φ((r z 2 1/2 (3.5

13 Boundary Harnack Principle and Martin Boundary at Infinity Proof We rewrite the Poisson kernel K B(0,r c as follows: K B(0,r c(x, z = G B(0,r c(x, y j( y z dy B(0,r c ( = + A(0,r,2qr { x z 2 x y } G B(0,r c(x, y j( y z dy =: I 1 + I 2 + I 3. + A(0,r,2qr { x z >2 x y } B(0,2qr c Note that, since x A(0, pr, qr and z B(0, r, fory A(0, r, 2qr with x z 2 x y, we have x y 1 2 ( x z 1 2 (p 1r >(4q 1 (p 1δ B(0,r c(y. (3.6 We claim that when y A(0, r, 2qr satisfies x z 2 x y, φ(r 2 1/2 G B(0,r c(x, y c 1 φ(δ B(0,r c(y 2 g(r c 2φ(r 2 1/2 φ(δ 1/2 B(0,r c(y 2 1/2 r d. (3.7 Since G B(0,r c(x, y g( x y, byeqs.2.9 and 3.6, we only need to prove the first inequality in Eq. 3.7 for (x, y satisfying y A(0, r,(p + 7r/8 and x z 2 x y. In this case, we have Let y 1 := (8 1 (p 1 + 1r y 1 y.then Thus by Eq. 3.8 x y 1 2 (p 1r > 4δ B(0,r c(y. (3.8 y 1 y δ B(0,r c(y δ B(0,r c(y (p 1r 1 x y. 4 x y 1 x y y 1 y 3 4 x y 3 (p 1r. (3.9 8 Because of Eq. 3.8, we can apply Theorem 2.9 and then use Eq. 3.9 to get that for (x, y satisfying y A(0, r,(p + 7r/8 and x z 2 x y, φ(δ B(0,r c(y 1 2 1/2 G B(0,r c(x, y c 3 φ(δ B(0,r c(y 2 G 1/2 B(0,r c(x, y 1 c 3 φ((64(p 1 2 r 2 1/2 φ(δ B(0,r c(y 2 1/2 g( x y 1 φ(r 2 1/2 c 4 φ(δ B(0,r c(y 2 g( 3 1/2 8 (p 1r c φ(r 2 1/2 5 g(r φ(δ B(0,r c(y 2 1/2 with constants c i = c i (φ, p, q >0, i = 3, 4, 5. In the last inequality we have used Eq Therefore using Eq. 2.9 we have proved Eq Applying Eq. 3.7 to I 1 and using the fact that δ B(0,r c(y y z, we get I 1 c 2 r d φ(r 2 1/2 φ( y z 2 1/2 j( y z dy. A(0,r,2qr { x z >2 x y }

14 P. Kim et al. Since B(0, r c B(z, r z c,byeq.2.8, the integral above is less than or equal to B(z,r z c φ ( y z 2 1/2 j( y z dy c6 r z φ(t 2 1/2 dt c 7 φ ( (r z 2 1/2, t where in the last inequality we used Eq Hence, I 1 c 8 φ(r 2 1/2 φ((r z 2 1/2 r d. To estimate I 2 we first note that if 2 x y x z,then y z x z y x 1 2 x z,hence,byeq.2.4, j( y z c 9 j( x z where c 9 = c 9 (φ > 0. Thus, I 2 c 10 j( x z g( x y dy c 11 j( x z B(x, x z 2 c 12 j( x z φ( x z 2 1 c 13 x z d c 14 r d. x z 2 0 t 1 φ(t 2 1 dt In the penultimate inequality, we used Eq Finally, we deal with I 3. For y 2qr and z < r we have that y z y z > y r (1 1/(2q y, hencebyeq.2.10, we get j( y z c 15 j( y. Also,for x < qr and y 2qr, we have that y x y x y /2, henceg( x y c 16 g( y by Eq Therefore, by Theorem 2.3, I 3 c 17 1 φ( y 2 dy c B(0,2qr c y d φ( y 2 y d 18 2qr t d 1 dt = c 19 r d. This concludes the proof of the Lemma. It is easy to see that, by the strong Markov property, for all Greenian open sets U and D with U D, G D (x, y = G U (x, y + E x GD (X τu, y for every (x, y R d R d. Thus, for all Greenian open sets U and D with U D, K D (x, z = K U (x, z + E x KD (X τu, z, (x, z U D c. (3.10 Since X is a purely discontinuous rotationally invariant Lévy process, it follows from 17, Proposition 4.1 (see also 22, Theorem 1 that if V is a Lipschitz open set and U V, P x (X τu V = 0 and P x (X τu dz = K U (x, zdz on V c. (3.11 Lemma 3.3 Let 1 < p < q <. There exists c = c(φ, p, q >1 such that for all r 1/2 andallopensetsu B(0, r c it holds that K U (x, y cr d ( U B(0, 1+p 2 r K U (z, y dz+1, for all x A(0, pr, qr U, y B(0, r. (3.12

15 Boundary Harnack Principle and Martin Boundary at Infinity Proof Let q 1 = 3+p 4, q 2 = 1+p (so that 1 < q 2 1 < q 2 < p ands q 1 r, q 2 r. Then, by Eqs and 3.11,forx A(0, pr, qr U and y B(0, r it holds that K U (x, y = E x K U (X τu B(0,s c, y + K U B(0,s c(x, y = K U (z, yk U B(0,s c(x, z dz + K U B(0,s c(x, y U B(0,s U B(0,q 2r Hence by Fubini s theorem, (q 2 q 1 rk U (x, y = 1 { z <s} K U (z, yk B(0,s c(x, z dz + K B(0,s c(x, y. q2r q 1r K U (x, y ds U B(0,q 2r q2r + q 1r ( q2r z K B(0,s c(x, z ds K U (z, y dz K B(0,s c(x, y ds =: I 1 + I 2. For s q 1 r, q 2 r and z U B(0, s, we have r < z < s q 2 r = 1+p 2 r < pr < x qr (q/q 1 s, so it follows from Lemmas 2.1 and 3.2 that K B(0,s c(x, z c 1 s d φ(s 2 1/2 φ ( (s z 2 1/2 c 2 r d φ ( (q 2 r 2 1/2 φ ( (s z 2 1/2 c 3 r d φ(r 2 1/2 φ ( (s z 2 1/2, where c 2 = c 2 (φ, p, q and c 3 = c 3 (φ, p, q.hence, q2r z q2r z K B(0,s c(x, z ds c 3 r d φ(r 2 1/2 φ(t 2 1/2 dt 0 c 4 r d φ(r 2 1/2 (q 2 r z φ ( (q 2 r z 2 1/2, where the last inequality follows from Eq Since t tφ(t 2 1/2 is increasing by Eq. 2.1 and q 2 r z q 2 r, we have that (q 2 r z φ((q 2 r z 2 1/2 q 2 rφ((q 2 r 2 1/2 c 5 rφ(r 2 1/2. Therefore, q2r z K B(0,s c(x, z ds c 6 r d+1. Further, for s q 1 r, q 2 r and y B(0, r we have y < r < q 1 r < s, andsos y > (q 1 1r, and we get similarly as above (but easier that I 2 = q2r q 1r K B(0,s c(x, y ds c 7 r d+1.

16 P. Kim et al. Finally, 1 K U (x, y = (q 2 q 1 r (I 1 + I 2 p 1 ( c 8 r d 4 U B(0,q 2r K U (z, y dz + 1, proving the lemma. Lemma 3.4 For every a > 1 there exists c = c(φ, a >1 such that for all r 1 and all open sets U B(0, r c it holds that ( 1 c K U(x, 0 K U (y, z dy + 1 U B(0,ar ( K U (x, z ck U (x, 0 K U (y, z dy + 1 (3.13 for all x U B(0, ar c and z B(0, r. U B(0,ar Proof Fix two constant b 2 and b 3 such that 1 > b 2 > b 3 > 1/a.LetU 1 := B(0, ar c U, U 2 := B(0, b 2 ar c U and U 3 := B(0, b 3 ar c U. Then by Eq. 3.10, forx U 1 and z B(0, r, K U (x, z = E x KU (X τu2, z + K U2 (x, z = K U (y, zp x (X τu2 dy + K U (y, zk U2 (x, y dy + K U2 (x, z U 3\U 2 U\U 3 =: I 1 + I 2 + I 3. We first estimate I 3 = K U2 (x, z = U 2 G U2 (x, y j( y z dy. For y U 2 we have that y > b 2 ar, hence ( 1 1 ( y y z y z y + z y. b 2 a b 2 a Hence, by Eq. 2.10, there exists c 1 = c 1 (φ > 0 such that c 1 1 j( y j( y z c 1 j( y. Therefore, c 1 1 K U 2 (x, 0 = c 1 1 G U2 (x, y j( y dy U 2 G U2 (x, y j( y z dy (3.14 U 2 = K U2 (x, z c 1 K U2 (x, 0 c 1 K U (x, 0. (3.15 In order to estimate I 2 = U\U 3 K U (y, zk U2 (x, y dy, we proceed similarly by estimating K U2 (x, y = U 2 G U2 (x,wj( w y dw for x U B(0, ar c and y U \ U 3. Note that since y / U 3 it holds that y < b 3 ar. For w U 2 it holds that w > b 2 ar. Hence, similarly as above we get that (1 b b 3 2 w w y (1 + b b 3 2 w. Thus there exists c 2 = c 2 (φ > 0 such that c 1 2 j( w j( w y c 2 j( w. Inthesame

17 Boundary Harnack Principle and Martin Boundary at Infinity way as above, this implies that c 1 2 K U 2 (x, 0 K U2 (x, y c 2 K U2 (x, 0 c 2 K U (x, 0. Therefore c 1 2 K U 2 (x, 0 K U (y, z dy I 2 (3.16 U\U 3 c 2 K U2 (x, 0 K U (y, z dy c 2 K U (x, 0 K U (y, z dy. (3.17 U\U 3 U B(0,ar In the case of I 1 we only need an upper estimate. It holds that I 1 = K U (y, zp x (X τu2 dy U 3\U 2 ( sup K U (y, z y U 3\U 2 P x (X τu2 B(0, b 2 ar. (3.18 By Lemma 3.3 (with p = b 3 a and q = b 2 a, there is a constant c 3 = c 3 (φ, a >0 such that ( ( sup K U (y, z c 3 r d K U (y, z dy + 1. y U 3\U 2 U\U 3 By Lemma 3.1 used with D = U 2, b 2 ar instead of r and 1 b 2 instead of a, thereisa constant c 4 = c 4 (φ, a >0 such that P x (X τu2 B(0, b 2 ar c 4 r d K U2 (x, 0. By applying the last two estimates to Eq we get ( I 1 c 5 K U2 (x, 0 K U (y, z dy + 1. (3.19 U\U 3 Putting together Eqs. 3.15, 3.17 and 3.19, we see that ( K U (x, z c 6 K U2 (x, 0 K U (y, z dy + 1 (3.20 U\U 3 ( c 6 K U (x, 0 K U (y, z dy + 1. U B(0,ar Thus, the upper bound in Eq holds true. In order to prove the lower bound, we may neglect I 1. First we note that for z B(0, r, K U (y, z dy = K U (y, z dy + K U (y, z dy U B(0,ar U\U 3 U 3\U 1 ( U3 K U (y, z dy + sup K U (y, z \ U 1 U\U 3 y U 3\U 1 ( K U (y, z dy + c 3 r d K U (y, z dy + 1 c 7 r d U\U 3 U\U 3 c 8 K U (y, z dy + 1. (3.21 (U\U 3

18 P. Kim et al. Here we used Lemma 3.3 in the second inequality. Next, by using the already proved upper bound (Eq. 3.20withz = 0, we see that ( K U (x, 0 c 6 K U2 (x, 0 K U (y, 0 dy + 1 U\U 3 ( c 6 K U2 (x, 0 K B(0, r c(y, 0 dy A(0,r,3r/2 ( c 6 K U2 (x, 0 c 9 r d dr + 1 A(0,r,3r/2 c 9 K U2 (x, 0. (3.22 Here we have used Lemma 3.2 in the third inequality. The lower estimate now follows from Eqs. 3.14, 3.16, 3.21 and Proof of Theorem 1.1 Let x U B(0, ar c. Then, by Eq. 3.11, u(x = B(0,r K U (x, 0 K U (x, zu(z dz B(0,r ( U B(0,ar ( = K U (x, 0 u(z dz + B(0,r ( = K U (x, 0 u(z dz + B(0,r K U (x, 0 u(z dz, B(0,ar K U (y, z dy + 1 u(z dz U B(0,ar U B(0,ar ( B(0,r u(y dy K U (y, zu(z dz dy where in the second line we used Lemma 3.4 and in the first, fourth and last line we used the fact that u vanishes a.e. on B(0, r c \ U. Proof of Corollary 1.2 It follows from Eq. 1.4 that for x, y U B(0, ar c, u(x v(x C 1K U (x, 0 B(0,ar u(z dz C 1 1 K U(x, 0 B(0,ar v(z dz = C2 B(0,ar u(z dz 1. B(0,ar v(z dz Similarly, u(y v(y C 1 1 K U(y, 0 B(0,ar u(z dz C 1 K U (y, 0 = C 2 1 B(0,ar v(z dz B(0,ar B(0,ar u(z dz. v(z dz The last two displays show that Eq. 1.5 is true for x, y U B(0, ar c with C 2 = C 4 1.

19 Boundary Harnack Principle and Martin Boundary at Infinity Corollary 3.5 For every a > 1, there exists c = c(d,φ,a >1 such that (i for every r 1,everyopensetU B(0, r c and every nonnegative function u on R d which is regular harmonic in U and vanishes a.e on B(0, r c \ U, u(x K U (x, 0 c u(y K U (y, 0, for all x, y U B(0, arc ; (ii for every r 1 and every nonnegative function u on R d which is regular harmonic in B(0, r c, u(x K B(0,r c(x, 0 c u(y K B(0,r c(y, 0, for all x, y B(0, arc. Proof The first claim is a direct consequence of Theorem 1.1 with c = C1 2, while the second follows from the first and the fact that the zero boundary condition is vacuous. Lemma 3.6 For every a > 1, there exists c = c(d,φ,a >1 such that for all r > 0 c 1 r d φ(r 2 G(x, 0 P x (τ B(0,r c < cr d φ(r 2 G(x, 0, for all x B(0, ar c. Proof Let x B(0, ar c. By the strong Markov property, G(x, y dy = E x G(X τb(0,r c, y dy,τ B(0,r c <. (3.23 B(0,r B(0,r Since r g(r is decreasing, 3, Lemma 5.53 shows that there exists a constant c = c(d such that for every r > 0 and all z B(0, r we have c g( y dy G(z, y dy g( y dy. B(0,r B(0,r B(0,r Then it follows from Eq that for x B(0, ar c, ( G(x, y dy g( y dy P x (τ B(0,r c <, B(0,r B(0,r with a constant depending on d only. By the uniform Harnack inequality (Theorem 2.7, there exists c 2 > 1 such that c 1 2 G(x, 0 G(x, y c 2G(x, 0 for every x B(0, ar c and y B(0, r.hence ( r d G(x, 0 g( y dy P x (τ B(0,r c <, x B(0, ar c, (3.24 B(0,r with a constant depending on d and a. It follows from Eqs. 2.9 and 2.5 that g( y dy 1 B(0,r φ(r 2, (3.25

20 P. Kim et al. with a constant depending on d and a. Combining Eqs we have proved the Lemma. Corollary 3.7 For every a > 1, there exists c = c(φ, d, a >1 such that for all r 1 it holds that c 1 φ(r 2 G(x, 0 K B(0,r c(x, 0 cφ(r 2 G(x, 0, x B(0, ar c, (3.26 and consequently lim K B(0,r c(x, 0 = 0. x Proof Note that P x (τ B(0,r c < = B(0,r K B(0,r c(x, z dz. Further, for z B(0, r and y B(0, r c we have that y z 2 y and hence j( y z c 1 j( y by Eq Therefore, K B(0,r c(x, z c 1 B(0,r c G B(0,r c(x, y j( y dy = c 1 K B(0,r c(x, 0. Using Eq it follows that P x (τ B(0,r c < c 1 B(0,r K B(0,r c(x, 0 dz = c 2 r d K B(0,r c(x, 0. On the other hand, from Lemma 3.1 with D = B(0, r c, we see that P x (τ B(0,r c < c 2 r d K B(0,r c(x, 0. Thus P x (τ B(0,r c < r d K B(0,r c(x, 0, x B(0, ar c. Comparing with the result in Lemma 3.6 gives Eq The last statement follows from the fact that lim x G(x, 0 = 0. Corollary 3.8 Let r 1 and U B(0, r c. If u is a non-negative function on R d which is regular harmonic in U and vanishes a.e. on B(0, r c \ U, then lim u(x = 0. x Proof Note that K U (x, 0 K B(0,r c(x, 0. It follows from Corollary 3.7 that lim K B(0,r c(x, 0 = 0. x Then the claim follows from Theorem 1.1. Remark 3.9 (i (ii Corollary 3.8 is not true if regular harmonic is replaced by harmonic. Indeed, let V denote the renewal function of the ladder height process of the onedimensional subordinate Brownian motion W d (S t. Then the function w(x = w( x, x d := V((x d + is harmonic in the upper half-space H B(( 0, 1, 1 c (see 10, vanishes on B(( 0, 1, 1 c \ H, but clearly lim xd w(x =. When d = 1 α, Corollary 3.8 is not true even for the symmetric α-stable process because the Green function of the complement of any bounded interval does not vanish at infinity, which can be seen using the Kelvin transform.

21 Boundary Harnack Principle and Martin Boundary at Infinity 4 The Infinite Part of the Martin Boundary In this section we will consider a large class of unbounded open sets D and identify the infinite part of the Martin boundary of D without assuming that the finite part of the Martin boundary of D coincides with the Euclidean boundary. We first recall the definition of κ-fatness at infinity from the introduction: Let κ (0, 1/2.AnopensetD in R d is κ-fat at infinity if there exists R > 0 such that for every r R, there exists A r R d such that B(A r,κr D B(0, r c and A r < κ 1 r. The origin does not play any special role in this definition: Suppose that D is κ-fat at infinity with characteristics (R,κ. For every Q R d,definer Q := R Q. For all r R Q,with r := A 2r and κ := κ/3, we have B( r, κr B(A 2r, 2κr D B(0, 2r c D B(0, r + Q c D B(Q, r c and  r Q A 2r + Q (κ/2 1 r + r <( κ 1 r. Remark 4.1 (i Note that it follows from the definition that any open set which is κ-fat at infinity is necessarily unbounded. (ii Since B(A r,κr B(0, r c we have that A r κr > r implying (κ + 1r < A r <κ 1 r. (iii We further note that B(A r,(κ/2r B(A (κ/2 1 r, r =. Indeed, for any point x in the intersection we would have that x A r +(κ/2r <(κ 1 + κ/2r, and at the same time x A (κ/2 1 r r >(κ+ 1(κ/2 1 r κr = (2κ κr. But this is impossible. In this section we first identify the infinite part of the Martin boundary of an open set D R d which is κ-fat at infinity with characteristics (R,κ. Without loss of generality, we assume that R > 1. In this section, the dependence of the lower case constants on κ may not be mentioned explicitly. Recall that we assume that (H1 and (H2 are true and d > 2(δ 2 δ 4. Lemma 4.2 Let D R d be an open set which is κ-fat at inf inity with characteristics (R,κ.Thereexistc= c(d,φ,κ>0 and γ = γ(d,φ,κ (0, d such that for every r R and any non-negative function h in R d whichisharmonicind B(0, r c it holds that h(a r c(κ/2 (d γk h(a (κ/2 k r, k = 0, 1, 2,.... (4.1 Proof Fix r R. For n = 0, 1, 2,..., let η n = (κ/2 n r, A n = A ηn and B n = B(A n,η n 1 (where η 1 = (κ/2r. Note that the balls B n are pairwise disjoint (cf. Remark 4.1 (iii. By harmonicity of h, for every n = 0, 1, 2..., h(a n =E An h(xτbn n 1 n 1 E An h(xτbn : X τbn B l = K Bn (A n, zh(z dz. B l l=0 l=0

22 P. Kim et al. By the uniform Harnack inequality, Theorem 2.7, there exists c 1 = c 1 (d,κ,φ>0 such that for every l = 0, 1, 2,..., h(z c 1 h(a l for all z B l.hence K Bn (A n, zh(z dz c 1 h(a l K Bn (A n, z dz, 0 l n 1. B l B l By Eq we have B l K Bn (A n, zdz c 2 φ(η 2 n 1 B l j( 2(A n z dz, 0 l n 1. For z B l, l = 0, 1,, n 1, it holds that z κ 1 (κ/2 l r + (κ/2 (l 1 r = (κ/2 l r(κ 1 + κ/2. Since A n κ 1 η n, we have that A n z A n + z 2κ 1 η n. Together with Theorem 2.3 and Lemma 2.1, this implies that j( 2(A n z c 3 j( η n for every z B l and 0 l n 1. Therefore, K Bn (A n, z dz c 4 j( η n φ(ηn 2 1 B l c 5 η d η n ηd l d l = c 5, 0 l n 1. B l ηn d Hence, ηn d h(a n 1 n c 5 ηl d h(a l, for all n = 1, 2,... l=0 Let a n := ηn dh(a n so that a n c n 1 5 l=0 a l. Using the identity 1 + c n 2 5 l=0 (1 + c 5 l = (1 + c 5 n 1 for n 3, by induction it follows that a n c 5 (1 + c 5 n 1 a 0. Let γ := log (1 + c 5 / log(2/κ so that (1 + c 5 n = (2/κ γ n. Note that c 5 can be chosen arbitrarily close to zero (but positive, so that γ<d. Thus, a 0 (1 + c 5 c 1 5 (κ/2γ n a n, or η0 dh(a 0 (1 + c 5 c 1 5 (κ/2γ n ηn dh(a n. Hence, h(a r (1 + c 5 c 1 5 (κ/2γ n (κ/2 dn h(a (κ/2 n r. Lemma 4.3 Let D R d be an open set which is κ-fat at inf inity with characteristics (R,κ. There exists c = c(d,φ,κ>0 such that for every r R and every non-negative function h on R d which is regular harmonic in D B(0,(κ/2 + 1r c, it holds that h(a r cr d h(z dz. B(0,r Proof Since h is regular harmonic in D B(0,(κ/2 + 1r c B(0,(κ/2 + 1r c, we have By Eq we have and B(A, κr 2 D ( h(a r = E Ar h X τb(ar, κr K 2 B(Ar, κr 2 (A r, zh(z dz. (4.2 B(0,r K B(Ar, κr 2 (A r, z c 1 j( 2(A r z φ ( ( κr 2 1, z B(0, r. (4.3 2

23 Boundary Harnack Principle and Martin Boundary at Infinity Since for z B(0, r we have that A r z <(κ 1 + 1r, by Eq we have j( 2(A r z c 2 j(r for some constant c 2 = c 2 (φ, κ > 0. Hence, combining Eqs and applying Lemma 2.1, we get h(a r c 3 j(rφ(r 2 1 h(z dz c 4 r d h(z dz B(0,r B(0,r which finishes the proof. Corollary 4.4 Let D R d be an open set which is κ-fat at inf inity with characteristics (R,κ. There exists c = c(d,φ,κ>0 such that for every r R with D B(0, r = and every w D B(0, r it holds that G D (A r,w cr d G D (z,wdz. (4.4 B(0,r Proof Let h( := G D (,w.thenh is regular harmonic in D B(0,(κ/2 + 1r c so the claim follows from Lemma 4.3. Lemma 4.5 Let D R d be an open set which is κ-fat at inf inity with characteristics (R,κ. For r > 0 and n = 0, 1, 2,...,letB n (r = B(0,(κ/2 n r. Thereexistc 1 = c 1 (d,φ,κ>0 and c 2 = c 2 (d,φ,κ (0, 1 such that for any r R and any nonnegative function h which is regular harmonic in D B(0, r c and vanishes in D c B(0, r c we have E x h ( X τd Bn(r c : X τd Bn(r c B(0, r c 1 c n 2 h(x, x D B n(r c, n = 0, 1, 2,... (4.5 Proof We fix r R. For n = 0, 1, 2,..., let B n = B n (r, B n = B n (r and η n = (κ/2 n r, and define h n (x := E x h (X τd Bcn : X τd B c B 0, x D B c n n. Then for x D B c n+1 we have h n+1 (x = E x h (X τd Bcn : τ c D B n+1 = τ c D B, X τd B c B 0 h n (x. n n Let A n = A ηn.then h n (A n = E An h (X τd Bcn : X τd B c B 0 E An h (X τbcn : X τb c B 0 n n = K c B (A n, zh(z dz. n B 0 By Lemma 3.2, there exists c 1 = c 1 (φ, κ > 0 such that for all z B 0 and n 1, ( K c B (A n, z c 1 A n z d( φ(η 2 n n 1/2 φ((η n z 2 1/2 + ηn d.

24 P. Kim et al. For z B 0 and n 1 we have that A n z η n and η n z η n, thus K c B (A n, z c 2 η d n n = c 2 (κ/2 nd r d, z B 0, n = 1, 2, Therefore, by Lemma 4.3 in the second inequality below and Lemma 4.2 in the third, we get that for n = 1, 2, 3..., h n (A n c 2 (κ/2 nd r d B 0 h(z dz c 3 (κ/2 nd h(a 0 c 4 (κ/2 γ n h(a n, where γ (0, d is the constant from Lemma 4.2. Now note that both h n 1 and h are regular harmonic in D B c n 1 and vanish on Bc n 1 Dc = B c n 1 \ D Bc n 1.Hence, h n (x h(x h n 1(x h(x h n 1 (A n 1 C 2 c 4 C 2 (κ/2 γ(n 1, x D B c n n=2, 3, 4..., h(a n 1 where the second inequality follows from Corollary 1.2. The cases n = 0 and n = 1 are clear by the harmonicity of h. Corollary 4.6 Let D R d be an open set which is κ-fat at inf inity with characteristics (R,κ. For r > 0 and n = 0, 1, 2,...,letB n (r = B(0,(κ/2 n r. Thereexistc 1 = c 1 (d,φ,κ>0and c 2 = c 2 (d,φ,κ (0, 1 such that for any r R with D B(0, r =, anyw D B(0, r and n 0, we have E x G D (X τd Bn(r c,w : X τd Bn(r c B(0, r c 1 c n 2 G D(x,w, x D B n (r c. The following Lemma is an analog of 2, Lemma 16 for infinity. The proof is essentially the same instead of using the balls that shrink to a finite boundary point, we use the complements of concentric balls with larger and larger radius (so they shrink at infinity. Lemmas 13 and 14 from 2 are replaced by our Corollary 1.2 and Lemma 4.5 respectively. Below we only indicate essential changes in the proof and refer the reader to the proof of 2, Lemma 16. Lemma 4.7 Let D R d be an open set which is κ-fat at inf inity with characteristics (R,κ.Thereexistc= c(d,φ,κ>0 and ν = ν(d,φ,κ>0 such that for any r R and all non-negative functions u and v on R d which are regular harmonic in D B(0, r/2 c, vanish in D c B(0, r/2 c and satisfy u(a r = v(a r, there exists the limit I(u,v= lim x, x D u(x v(x, and we have ( u(x v(x I(u,v x ν c, x D B(0, r c. (4.6 r Proof Let r R be fixed. Without loss of generality assume that u(a r = v(a r = 1. Let n 0 (d,φ N to be chosen later, and let a = (κ/2 n0. For n = 0, 1, 2,...,define r n = a n r, B c n = B(0, r n c, D c n = D Bc n, n = D c n \ Dc n+1, 1 = B(0, r.

25 Boundary Harnack Principle and Martin Boundary at Infinity For l = 1, 0, 1,...,n 1 let u l n (x := E x v l n (x := E x u (X τdcn : X τd c l, n x R d, (4.7 v (X τdcn : X τd c l, n x R d. (4.8 Note that since l B(0, r l+1, it holds that u l n (x E x u (X τdcn : X τd c B(0, r l+1. n Denote the constants c 1 and c 2 in Lemma 4.5 by C and ξ respectively. Apply Lemma 4.5 with r = r l+1.thenr n = (κ/2 n0n r = (κ/2 n0(n l 1 r, henceforn = 0, 1, 2,... and x D c n, u l n (x C(ξ n0 n l 1 u(x, l = 1, 0, 1,...,n 2. Choose n 0 large enough so that (1 ξ n Then since n 2 l= 1 (ξ n0 n l 1 = ξ n0 n 1 k=0 (ξ n0 k ξ n0 (1 ξ n0 1 2ξ n0, we have that for n = 1, 2,... and x D c n, n 2 l= 1 u l n (x 2 Cξ n0 u(x. Let ɛ be a number in (0, 1. We can redefine n 0 (ɛ, d,φ so that for n = 1, 2,..., l = 1, 0, 1,...,n 2, u l n (x ɛn 1 l u n 1 n (x, x D c n. By symmetry we can also achieve that for n = 1, 2,..., l = 1, 0, 1,..., n 2, vn l (x ɛn 1 l vn n 1 (x, x D c n. Now we claim that there exist constants c 1 = c 1 (d,φ,κ>0 and ζ = ζ(d,φ,κ (0, 1 such that for all l = 0, 1,..., sup x D c l u(x v(x (1 + c 1ζ l u(x inf x D c v(x. l From now on the proof is essentially the same as the proof of 2, Lemma 16, hence we omit it. Remark 4.8 (i Assume that u and v are nonnegative functions on R d which are regular harmonic in D B(0, r/2 c and vanish in D c B(0, r/2 c.defineũ r and ṽ r by ũ r (x := u(x u(a r, ṽ r(x := v(x v(a r. Then ũ r and ṽ r satisfy the assumptions of Lemma 4.7, in particular ũ r (A r = ṽ r (A r. Hence, there exists the limit I(ũ r, ṽ r = lim x, x D ũ r (x ṽ r (x.

26 P. Kim et al. Therefore we can conclude that there exists the limit I(u,v,A r = u(x lim x, x D v(x = u(a r v(a r I (ũ r, ṽ r. Suppose that ρ R is another radius such that u and v are regular harmonic in D B(0,ρ/2 c and vanish in D c B(0,ρ/2 c. Then the same argument using A ρ instead of A r would give that there exists the limit I(u,v,A ρ = u(x lim x, x D v(x = u(a ρ v(a ρ I(ũ ρ, ṽ ρ. This shows that the limit is independent of the point A r. (ii It easily follows from Eq. 4.6 that there exist c = c(d,φ,κ> 0 and ν = ν(d,φ,κ>0 such that for any r R, u(x v(x u(y v(y c x y ν r x, y D B(0, r c for all non-negative functions u and v on R d which are regular harmonic in D B(0, r/2 c, vanish in D c B(0, r/2 c and satisfy u(a r = v(a r. From now on D will be an open set which is κ-fat at infinity with characteristics (R,κ.Fixx 0 D B(0, R c and recall that M D (x, y = G D(x, y G D (x 0, y, x, y D, y = x 0. For r > 2( x x 0, both functions y G D (x, y and y G D (x 0, y are regular harmonic in D B(0, r/2 c and vanish on D c B(0, r/2 c.hence,asanimmediate consequence of Lemma 4.7 and Remark 4.8 (i we get the following theorem. Theorem 4.9 For each x D there exists the limit M D (x, := lim M D(x, y. y D, y Recall that X D is the process X killed upon exiting D. As the process X D satisfies Hypothesis (B in 15, D has a Martin boundary M D with respect to X satisfying the following properties: (M1 (M2 (M3 D M D is a compact metric space (with the metric denoted by d; D is open and dense in D M D, and its relative topology coincides with its original topology; M D (x, can be uniquely extended to M D in such a way that (a M D (x, y converges to M D (x,w as y w M D in the Martin topology, (b for each w D M D the function x M D (x,w is excessive with respect to X D,

27 Boundary Harnack Principle and Martin Boundary at Infinity (c the function (x,w M D (x,w is jointly continuous on D (D M D in the Martin topology and (d M D (,w 1 = M D (,w 2 if w 1 = w 2 and w 1,w 2 M D. In the remainder of the paper whenever we speak of a bounded or an unbounded sequence of points we always mean in the Euclidean metric (and not in the Martin metric d. Definition 4.10 A point w M D is called a finite Martin boundary point if there exists a bounded sequence (y n n 1, y n D, converging to w in the Martin topology. A point w M D is called an infinite Martin boundary point if every sequence (y n n 1, y n D, converging to w in the Martin topology is unbounded. The set of finite Martin boundary points is denoted by f M D, and the set of infinite Martin boundary points by M D. Remark 4.11 Suppose that w f M D and let (y n n 1 D be a bounded sequence converging to w in the Martin topology. Then (y n n 1 has a subsequence (y nk k 1 converging to a point y in the Euclidean topology. It cannot happen that y D, because in this case we would have that lim ynk y M D (x, y nk = M D (x, y implying by (M3(d that y = w. Therefore, y D the Euclidean boundary of D.Inparticular, this shows that for every ɛ>0, the sequence (y n n 1 (converging to w f M D in the Martin topology can be chosen so that δ D (y n <ɛfor all n 1. Proposition 4.12 Let D be an open set which is κ-fat at inf inity. Then M D consists of exactly one point. Proof Let w M D and let M D(,w be the corresponding Martin kernel. If the sequence (y n n 1 D converges to w in the Martin topology, then, by (M3(a, M D (x, y n converge to M D (x,w. On the other hand, y n, thus by Theorem 4.9, lim M D(x, y n = lim M D(x, y n = M D (x,. n y n Hence, for each w M D it holds that M D(,w= M D (,. Since, by (M3(d, for two different Martin boundary points w (1 and w (2 it always holds that M D (,w (1 = M D (,w (2, we conclude that the infinite part of the Martin boundary can be identified with the single point. From now on we use the notation M D ={ } and, for simplicity, we sometimes continue to write M D (x, for the more precise M D (x,. We now briefly discuss some properties of the finite part of the Martin boundary. Recall that d denotes the Martin metric. For ɛ>0let { } K ɛ := w f M D : d(w, ɛ (4.9 be a closed subset of M D. By the definition of the finite part of the Martin boundary, for each w K ɛ there exists a bounded sequence (y w n n 1 D such that lim n d(y w n,w= 0. Without loss of generality we may assume that d(yw n,w< ɛ 2 for all n 1.